Mathematics - Quadratic Equation Question with Solution | TestHub

$$x^2-(a+1)x+a^2+a-8=0$$ Since roots are different, therefore $\mathrm{D}>0$ $$\begin{array}{rl}& \Rightarrow (a+1{)}^2-4\left(a^2+a-8\right)>0\\ & \Rightarrow (a-3)(3a+1)<0\end{array}$$ There are two cases arises. Case I. $a-3>0$ and $3a+1<0$ $\Rightarrow a>3$ and $a<-\frac{11}{3}$ Hence, no solution in this case Case II : $a-3<0$ and $3a+11>0$ $\Rightarrow a<3$ and $a>-\frac{11}{3}$ $\therefore -\frac{11}{3}<a<3\Rightarrow -2<a<3$